QPaperGen

We are given:
\[
f(x) = x + \frac{1}{x}
\]

Differentiate with respect to \( x \):
\[
f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2}
\]

Set \( f'(x) = 0 \):
\[
1 - \frac{1}{x^2} = 0 \Rightarrow \frac{1}{x^2} = 1 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1
\]

So, the critical points are \( x = -1 \) and \( x = 1 \), which divide the real line into intervals:
\[
(-\infty, -1),\quad (-1, 1),\quad (1, \infty)
\]

Now analyze the sign of \( f'(x) \) in each interval:


\[\] In \( (-\infty, -1) \):
\( x < -1 \Rightarrow x^2 > 1 \Rightarrow \frac{1}{x^2} < 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} > 0 \)
So, \( f \) is strictly increasing on \( (-\infty, -1) \).

\[\] In \( (-1, 1) \):
\( -1 < x < 1,\ x \neq 0 \Rightarrow x^2 < 1 \Rightarrow \frac{1}{x^2} > 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} < 0 \)
So, \( f \) is strictly decreasing on \( (-1, 1) \).

\[\] In \( (1, \infty) \):
\( x > 1 \Rightarrow x^2 > 1 \Rightarrow \frac{1}{x^2} < 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} > 0 \)
So, \( f \) is strictly increasing on \( (1, \infty) \).
\[\]

Hence, \( f(x) = x + \frac{1}{x} \) is strictly increasing on any interval \( \mathbf{I} \) such that \( \mathbf{I} \cap (-1, 1) = \emptyset \).
That is, \( f \) is strictly increasing on any interval disjoint from \( (-1, 1) \).